What is the derivative of (2x-1)^2(x^3+1)^2 ?

The derivative of (2x-1)^2(x^3+1)^2 is 32x^7-28x^6+6x^5+40x^4-32x^3+6x^2+8x-4

What is the first derivative of (2x-1)^2(x^3+1)^2 ?

The first derivative of (2x-1)^2(x^3+1)^2 is 32x^7-28x^6+6x^5+40x^4-32x^3+6x^2+8x-4

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derivative of

(2 x - 1)^{2} (x^{3} + 1)^{2}

32 x^{7} - 28 x^{6} + 6 x^{5} + 40 x^{4} - 32 x^{3} + 6 x^{2} + 8 x - 4

Solution steps

\frac{d}{d x} ((2 x - 1)^{2} (x^{3} + 1)^{2})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = (2 x - 1)^{2}, g = (x^{3} + 1)^{2}

= \frac{d}{d x} ((2 x - 1)^{2}) (x^{3} + 1)^{2} + \frac{d}{d x} ((x^{3} + 1)^{2}) (2 x - 1)^{2}

\frac{d}{d x} ((2 x - 1)^{2}) = 4 (2 x - 1)

\frac{d}{d x} ((x^{3} + 1)^{2}) = 6 x^{2} (x^{3} + 1)

= 4 (2 x - 1) (x^{3} + 1)^{2} + 6 x^{2} (x^{3} + 1) (2 x - 1)^{2}

Simplify

4 (2 x - 1) (x^{3} + 1)^{2} + 6 x^{2} (x^{3} + 1) (2 x - 1)^{2} : 32 x^{7} - 28 x^{6} + 6 x^{5} + 40 x^{4} - 32 x^{3} + 6 x^{2} + 8 x - 4

= 32 x^{7} - 28 x^{6} + 6 x^{5} + 40 x^{4} - 32 x^{3} + 6 x^{2} + 8 x - 4

y = cos (\frac{π t}{4} - \frac{π}{3})

\int \frac{2}{x ^{5}} d x

\int cot (u) d u

y^{^{'}} = 2 x (x - 2)

n = 0 \sum \infty \frac{1 ^{n}}{n + 2}

What is the derivative of (2x-1)^2(x^3+1)^2 ?
The derivative of (2x-1)^2(x^3+1)^2 is 32x^7-28x^6+6x^5+40x^4-32x^3+6x^2+8x-4
What is the first derivative of (2x-1)^2(x^3+1)^2 ?
The first derivative of (2x-1)^2(x^3+1)^2 is 32x^7-28x^6+6x^5+40x^4-32x^3+6x^2+8x-4